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c++functionrandom
Bobby
Bobby

Reputation: 37

Random Number Generator Within Given Range

I'm trying to write a program that uses a function to generate 10 random numbers within a range provided by the user. It seems to work okay, other than the fact that the numbers returned are all 1's:

#include <iostream>
#include <cstdlib>
#include <ctime>
using namespace std;

int rand_int(int min, int max);

int main()
{
    int min, max;

    cout << "Hello user.\n\n"
         << "This program will generate a list of 10 random numbers within a 
         given range.\n"
         << "Please enter a number for the low end of the range: ";
    cin  >> min;
    cout << "You entered " << min << ". \n"
         << "Now please enter a number for the high end of the range: ";
    cin  >> max;

    while(min > max){
        cout << "Error: Your low number is higher than your high number.\n"
             << "Please reenter your high number, or press ctrl + c 
                 to end program.\n";
        cin  >> max;
        cout << endl;
    }

    for(int i = 0; i < 10; i++){
        int rand_int(int min, int max);
        cout << rand_int << endl;
    }

    return 0;
}


int rand_int(int min, int max)
{
    srand(time(0)); // Ensures rand will generate different numbers at different times

    int range = max - min;

    int num = rand() % (range + min);

    return num;
}

Upvotes: 2

Views: 2921

Answers (3)

Yash
Yash

Reputation: 2023

The Fastest & Simplest way to get the Random Number within a range is-

int lower=1,upper=10;//for example
srand(time(0));
int y = (rand() % (upper-lower + 1)) + lower;

It will give you the output in the range- [1,10] (both inclusion).

That's it. Cheers!

Upvotes: 0

Shafik Yaghmour
Shafik Yaghmour

Reputation: 158459

Having warnings turned on may have helped here, with -Wall flag gcc tells us:

warning: the address of 'int rand_int(int, int)' will always evaluate as 'true' [-Waddress]
     cout << rand_int << endl;
             ^

Although clang gives a warning without the need to add flags. You are using a function pointer here and since std::cout does not have an overload for a function pointer it is selecting the bool overload and converting the function pointer to true. The calls should be like this:

std::cout << rand_int(min, max)  <<std::endl;

Although that will not totally fix your issues, you also need to move:

srand(time(0));

outside your function preferably at the start of your program. Since you are calling rand_int ten times very quickly the result of time(0) will probably be the same and therefore you will return the same 10 numbers.

This line:

int rand_int(int min, int max);

in the for loop is just a redeclaration of the function and is not needed.

Although, if C++11 is an option using the random header makes way more sense and is much simpler:

#include <iostream>
#include <random>

int main()
{
    std::random_device rd;

    std::mt19937 e2(rd());

    std::uniform_int_distribution<int> dist(1,10);

    for (int n = 0; n < 10; ++n) {
            std::cout << dist(e2) << ", " ;
    }
    std::cout << std::endl ;
}

If C++11 is not an option then your should at least check out the How can I get random integers in a certain range? C FAQ entry which gives the following formula for generating numbers in the range [M, N]:

M + rand() / (RAND_MAX / (N - M + 1) + 1)

and of course there is always boost:

#include <iostream>
#include <boost/random/mersenne_twister.hpp>
#include <boost/random/uniform_int_distribution.hpp>

int main()
{
  boost::random::mt19937 gen;
  boost::random::uniform_int_distribution<> dist(1, 10);

  for (int n = 0; n < 10; ++n) {
    std::cout << dist(gen) << ", ";
  }
  std::cout << std::endl ;
}

Upvotes: 4

rhughes
rhughes

Reputation: 9583

Try changing this:

for(int i = 0; i < 10; i++){
    int rand_int(int min, int max);
    cout << rand_int << endl;
}

to:

for(int i = 0; i < 10; i++){
    int myRandomNumber = rand_int(int min, int max);
    cout << myRandomNumber << endl;
}

It seems that you were outputting the function rather than the return result of it.

Upvotes: 0

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